Difference between revisions of "1958 AHSME Problems/Problem 46"
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From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. Substituting both of them in, you get that if <math>x=2</math>, then the value is <math>1</math>. If you plug in the value of <math>x=0</math>, you get the value of <math>-1</math>. So the answer is <math>\textbf{(D)}</math> | From there, we get that <math>x</math> is either <math>2</math> or <math>0</math>. Substituting both of them in, you get that if <math>x=2</math>, then the value is <math>1</math>. If you plug in the value of <math>x=0</math>, you get the value of <math>-1</math>. So the answer is <math>\textbf{(D)}</math> | ||
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+ | Solution by: the_referee | ||
== See Also == | == See Also == |
Revision as of 12:23, 23 October 2018
Problem
For values of less than
but greater than
, the expression
has:
Solution
From , we can further factor
and then
and finally
. Using
, we can see that
. From there, we can get that
.
From there, we get that is either
or
. Substituting both of them in, you get that if
, then the value is
. If you plug in the value of
, you get the value of
. So the answer is
Solution by: the_referee
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 45 |
Followed by Problem 47 | |
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